YES

Problem:
 f(s(x),y) -> f(x,s(x))
 f(x,s(y)) -> f(y,x)
 f(c(x),y) -> f(x,s(x))

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = 3x0 + 3,
   
   [f](x0, x1) = 4x0 + 2x1 + 3,
   
   [s](x0) = 4x0 + 6
  orientation:
   f(s(x),y) = 16x + 2y + 27 >= 12x + 15 = f(x,s(x))
   
   f(x,s(y)) = 4x + 8y + 15 >= 2x + 4y + 3 = f(y,x)
   
   f(c(x),y) = 12x + 2y + 15 >= 12x + 15 = f(x,s(x))
  problem:
   f(c(x),y) -> f(x,s(x))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 1 1]     [1]
    [c](x0) = [1 1 1]x0 + [1]
              [1 0 0]     [0],
    
                  [1 1 1]     [1 1 0]     [1]
    [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [0]
                  [0 0 0]     [0 0 0]     [0],
    
              [1 0 0]     [0]
    [s](x0) = [1 1 1]x0 + [1]
              [0 0 0]     [0]
   orientation:
                [3 2 2]    [1 1 0]    [3]    [3 2 2]    [2]            
    f(c(x),y) = [0 0 0]x + [0 0 0]y + [0] >= [0 0 0]x + [0] = f(x,s(x))
                [0 0 0]    [0 0 0]    [0]    [0 0 0]    [0]            
   problem:
    
   Qed